For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive.If p is the smallest prime factor of h(100) + 1, then p is
this is definitely a difficult number properties question. Let's first consider the prime factors of h(100). According to the given function, h(100) = 2*4*6*8*...*100
By factoring a 2 from each term of our function, h(100) can be rewritten as 2^50*(1*2*3*...*50).
Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).
Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.
Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.
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发表于 2008-9-3 17:47:00
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